435=(1+n)/2*n

Solution for 435=(1+n)/2*n equation:

n in (-oo:+oo)

435 = n*((n+1)/2) // - n*((n+1)/2)

435-(n*((n+1)/2)) = 0

(-1*n*(n+1))/2+435 = 0

(-1*n*(n+1))/2+(2*435)/2 = 0

2*435-1*n*(n+1) = 0

870-n^2-n = 0

870-n^2-n = 0

870-n^2-n = 0

DELTA = (-1)^2-(-1*4*870)

DELTA = 3481

DELTA > 0

n = (3481^(1/2)+1)/(-1*2) or n = (1-3481^(1/2))/(-1*2)

n = -30 or n = 29

(n+30)*(n-29) = 0

((n+30)*(n-29))/2 = 0

((n+30)*(n-29))/2 = 0 // * 2

(n+30)*(n-29) = 0

( n+30 )

n+30 = 0 // - 30

n = -30

( n-29 )

n-29 = 0 // + 29

n = 29

n in { -30, 29 }